Integration with Trigonometric Functions

Integration by Parts gives us $\int u dv = uv - \int vdu$ . We let $u = x^2$ and $dv = \cos x dx$ , so we have $du = 2x dx$ and $v = \sin x$ . Substituting these in, we have

$\displaystyle \int_0^{\frac{3\pi}{2}} x^2\cos x dx = [x^2\sin x]_0^{\frac{3\pi}{2}} - \int_0^{\frac{3\pi}{2}} 2x\sin x dx = -\frac{9\pi^2}{4} - 2\int_0^{\frac{3\pi}{2}} x\sin x dx$

We again use integration by parts, this time letting $u = x$ and $dv = \sin x$ , which gives $du = dx$ and $v = -\cos x dx$ . Substituting these into the above equation, we have

$\begin{aligned} \displaystyle -\frac{9\pi^2}{4} - 2\int_0^{\frac{3\pi}{2}} x\sin x dx &= -\frac{9\pi^2}{4} -2\left([-x\cos x]_0^{\frac{3\pi}{2}} - \int_0^{\frac{3\pi}{2}} (-\cos x)dx\right) \\ &= -\frac{9\pi^2}{4} -2 \int_0^{\frac{3\pi}{2}} \cos x dx \\ &= -\frac{9\pi^2}{4} + 2 \\ \end{aligned}$

Hence $a + b + c = 2 + 9 + 4 = 15$ .